Suppose the student finishes the exam at time t.
P(t<y)=y/3, 0≤y≤2 The 1conditional probability that the full time is used will be
P(t>2∣t>1.75)=P(t>1.75)P(t>2∩t>1.75)=
=P(t>1.75)P(t>2)=1−P(t≤1.75)1−P(t≤2)=
=1−1.75/31−2/3=0.8 b. There are 4 choices per question, and 7 questions. So there are
4×4×4×4×4×4×4=47=16384 ways to answer.
What is the probability that two papers have the same answer to the first question?
p=41 What is the probability that two papers have the same answers ?
P(same)=(41)7=163841≈0.000061
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